A+bi is a complex number
(3+2i)(a+bi)=17+7i
remember
i²=-1
so
expand and solve
a is the ral part
bi is the imaginary part
ok so
if Ac+Be=dc+fe where c=c and e=e then A=d and B=f
(3+2i)(a+bi)=17+7i
expand/distribute/FOIL
3a+2ai+3bi+2bi²=17+7i
3a+2bi+2ai+2bi=17+7i
3a-2b+2ai+2bi=17+7i
real parts are 3a-2b
imaginary is 2ai+2bi
so
3a-2b=17 and
2ai+2bi=7i
we need to solve
2nd equation, divide both sides by i
2a+2b=7
multiply by -1 and add to other equation
3a+2b=17
<u>-2a-2b=-7 +</u>
1a+0b=10
a=10
subsitue
2a+2b=7
2(10)+2b=7
20+2b=7
2b=-13
b=-13/2
the complex number is
10-(13/2)i or
10-6.5i
Answer:
41
Step-by-step explanation:
BIDMAS
[5 x (4 + 6) - 9]
Brackets first
4 + 6 = 10
[5 x (10) - 9]
Multiply next
5 x 10 = 50
Subtract last
50 - 9 = 41
9/-7 x -4=
36/7
Hope it helps
Answer:
a) The domain is the set of the possible values of x.
In the graph, we can see that x goes from - 4 to 4
So the domain is: D = x∈ {-4, 4)
b) The zeros are the points where the graph cuts the x-axis.
Those values are:
x = -2
x = 0
x = 2.
c) The graph is positive if it is above the x-axis and it is negative if it is below the x-axis.
in the interval {-4,0} the graph is bellow the x-axis, so in this interval the function is negative.
in the interval {0, 2} the graph is above the x-axis, so in this interval the function is positive.
in the interval {2, 4} the graph is bellow the x-axis, so in this interval the function is negative.
